# How do you solve #64x^3+27=0#?#?

##### 2 Answers

#### Answer:

Factor

Hence

#### Explanation:

In general

Putting

#64x^3+27 = (4x)^3+3^3 = (4x + 3)((4x)^2 - (4x)*3 + 3^2)#

#= (4x+3)(16x^2 - 12x + 9)#

The remaining quadratic factor has no linear factors with Real coefficients.

You can tell that because its discriminant is negative:

Its discriminant

#Delta = b^2 - 4ac = (-12)^2 - (4xx16xx9) = 144 - 576#

#= -432#

In fact,

where

#### Answer:

#### Explanation:

(As an alternate method)